Question: How to prove the following identity? $$ \sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m. $$ I'm also looking for the generalization of this identity like $$ \sum_{s=k}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=? $$
Proofs, hints, or references are all welcome.
Let $$a_m=\sum_{s=0}^{m}\binom{2s}{s}\binom{s}{m-s}\dfrac{(-1)^s}{s+1}$$and let $P(x)=\displaystyle\sum_{m=0}^{\infty}a_m x^m$ be the generating function for $a_m$. We can rewrite $P(x)$ as $$\sum_{m=0}^{\infty}\sum_{s=0}^{m}\binom{2s}{s}\binom{s}{m-s}\frac{(-1)^s}{s+1}x^m=\sum_{s=0}^{\infty}\sum_{m=s}^{\infty}\binom{2s}{s}\binom{s}{m-s}\frac{(-1)^s}{s+1}x^m$$The second sum is really the same thing as a sum of $m-s$ from $0$ to $\infty$, so we can write it as so and take out the terms that don't involve $m$: $$\sum_{s=0}^{\infty}\binom{2s}{s}\frac{(-1)^s}{s+1}x^s\sum_{m-s=0}^{\infty}\binom{s}{m-s}x^{m-s}$$By the binomial theorem, the above is equal to $$\sum_{s=0}^{\infty}\binom{2s}{s}\frac{(-1)^s}{s+1}x^s (1+x)^s=\sum_{s=0}^{\infty}\binom{2s}{s}\frac{1}{s+1}(-x-x^2)^s$$From the generating function of the Catalan numbers, $\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}\dfrac{1}{n+1}x^n=\dfrac{1-\sqrt{1-4x}}{2x}$, this last expression is equal to $$\frac{1-\sqrt{1+4x+4x^2}}{-2x-2x^2}=\frac{-2x}{-2x-2x^2}=\frac{1}{x+1}=1-x+x^2-x^3+\cdots$$So to conclude, we know that $$\sum_{m=0}^{\infty}a_m x^m=\sum_{m=0}^{\infty}(-1)^m x^m\Leftrightarrow a_m=(-1)^m$$This method is known as the snake oil method. To evaluate the more general expression, you'll need to truncate some of the terms at the beginning of the sum.
This is just a supplement to the nice answer of @tc2718. We show that it is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. We can write e.g. $$\binom{n}{k}=[x^k](1+x)^n$$
We obtain \begin{align*} \sum_{s=0}^{m}&{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}\\ &= \sum_{s=0}^{m}[u^{m-s}](1+u)^s(-1)^s[x^s]\frac{1-\sqrt{1-4x}}{2x}\tag{1}\\ &= [u^{m}]\sum_{s=0}^{m}(1+u)^s(-u)^s[x^s]\frac{1-\sqrt{1-4x}}{2x}\tag{2}\\ &= [u^{m}]\frac{1-\sqrt{1-4(1+u)(-u)}}{2(1+u)(-u)}\tag{3}\\ &= [u^{m}]\frac{1-(1+2u)}{2(1+u)(-u)}\\ &= [u^{m}]\frac{1}{1+u}\\ &= [u^{m}]\sum_{s=0}^{\infty}(-u)^s\\ &=(-1)^m \end{align*}
Comment:
In (1) we use the coefficient of operator together with the series expansion of the Catalan-numbers: $\frac{1-\sqrt{1-4x}}{2x}=\sum_{s=0}^{\infty}\frac{1}{s+1}\binom{2s}{s}x^s$
In (2) we use the rule $[x^s]f(x)=[x^0]x^{-s}f(x)$
In (3) we use the substitution rule $f(x):=\sum_{s=0}^{\infty}a_sx^s=\sum_{s=0}^\infty[y^s]f(y)x^s$
Suppose we seek to verify that
$$\sum_{s=0}^m {2s\choose s} {s\choose m-s} \frac{(-1)^s}{s+1} = (-1)^m$$
without using the generating function of the Catalan numbers.
Re-write the sum as follows: $$\sum_{s=0}^m {2s\choose m} {m\choose s} \frac{(-1)^s}{s+1}$$
which is $$\frac{1}{m+1} \sum_{s=0}^m {2s\choose m} {m+1\choose s+1} (-1)^s$$
which turns into $$- \frac{1}{m+1} \sum_{s=1}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s} \\ = \frac{1}{m+1} {-2\choose m} - \frac{1}{m+1} \sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s} \\ = \frac{1}{m+1} (-1)^m \frac{(m+1)!}{m!} - \frac{1}{m+1} \sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s} \\ = (-1)^m - \frac{1}{m+1} \sum_{s=0}^{m+1} {2s-2\choose m} {m+1\choose s} (-1)^{s}.$$
Now introduce $${2s-2\choose m} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{2s-2} \; dz.$$
We get for the sum $$- \frac{1}{m+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2} \sum_{s=0}^{m+1} {m+1\choose s} (-1)^{s} (1+z)^{2s} \; dz \\ = -\frac{1}{m+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2} (1-(1+z)^2)^{m+1}\; dz \\ = \frac{(-1)^m}{m+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+1}} \frac{1}{(1+z)^2} (z^2+2z)^{m+1} \; dz \\ = \frac{(-1)^m}{m+1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{(1+z)^2} (z+2)^{m+1} \; dz = 0.$$
This concludes the argument.
